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lys-0071 [83]
2 years ago
13

Rob had twice as much money as Sam. Then Sam gave Rob 1 quarter, 2 nickels, and 3 pennies. Rob then gave Sam 8 dimes. If they no

w have the same amount of money, how much money did Rob originally have? (Steps please!)
Mathematics
1 answer:
bezimeni [28]2 years ago
6 0

Let the initial amount that Sam had be x.

Then, since Rob had twice as much as Sam, Rob's money would be 2x.

Now, Sam gave 1 quarter, 2 nickels and 3 pennies to Rob.

Let's convert all of them into pennies.

1 quarter + 2 nickels + 3 pennies = 25 + (2 x 5) + 3

                                                      = 38 pennies

So, Sam's amount is x - 38 and Rob's amount is 2x + 38.

Now, Rob gives 8 dimes to Sam.

8 dimes = 80 pennies.

So, Sam's amount now is x - 38 + 80 = x + 42 and

Rob's amount is 2x + 38 - 80 = 2x - 42

Its given that these two amounts are same.

So,  2x - 42 = x + 42

2x - x = 42 + 42

x = 84

Rob's original amount = 2x = 2(84) = 168 pennies.

                                                         = $1 + 6 dimes + 1 nickel + 3 pennies

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The Michigan veteran population in 1990 was 14.2% and in 2000 was 12.4%.
Mandarinka [93]

Let x- intercept represents time

Let y-intercept represents population

Let 1990 represent initial year

Points are (0,14.2)(10,12.4)

slope m= \frac{y2-y1}{x2-x1}

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General linear equation

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plug the 1st point in equation

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2 years ago
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2 years ago
Find the 2th term of the expansion of (a-b)^4.​
vladimir1956 [14]

The second term of the expansion is -4a^3b.

Solution:

Given expression:

(a-b)^4

To find the second term of the expansion.

(a-b)^4

Using Binomial theorem,

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here, a = a and b = –b

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

Substitute i = 0, we get

$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4

Substitute i = 1, we get

$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b

Substitute i = 2, we get

$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}

Substitute i = 3, we get

$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}

Substitute i = 4, we get

$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}

Therefore,

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}

Hence the second term of the expansion is -4a^3b.

3 0
2 years ago
15% of $9 is $ this is a question on my math worksheet please respond quick and show the steps on how to do it please and thank
Vladimir79 [104]
9 times .15 = 1.35
15% of $9 is $1.35
5 0
2 years ago
Read 2 more answers
4. Compare: 63% and 0.8 *
vovangra [49]

Answer:

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Step-by-step explanation:

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3 years ago
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